Inverse task matrix. Higher mathematics

The inverse matrix for a given matrix is ​​such a matrix, multiplying the original one by which gives the identity matrix: A mandatory and sufficient condition for the presence inverse matrix is that the determinant of the original one is not equal to zero (which in turn implies that the matrix must be square). If the determinant of a matrix is ​​equal to zero, then it is called singular and such a matrix does not have an inverse. In higher mathematics, inverse matrices are important and are used to solve a number of problems. For example, on finding the inverse matrix a matrix method for solving systems of equations was constructed. Our service site allows calculate inverse matrix online two methods: the Gauss-Jordan method and using the matrix of algebraic additions. Interrupt implies a large number of elementary transformations inside the matrix, the second is the calculation of the determinant and algebraic complements to all elements. To calculate the determinant of a matrix online, you can use our other service - Calculation of the determinant of a matrix online

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This topic is one of the most hated among students. Worse, probably, are the qualifiers.

The trick is that the very concept of an inverse element (and I’m not just talking about matrices) refers us to the operation of multiplication. Even in school curriculum Multiplication is considered a complex operation, and multiplication of matrices is generally a separate topic, to which I have a whole paragraph and a video lesson dedicated.

Today we will not go into the details of matrix calculations. Let’s just remember: how matrices are designated, how they are multiplied, and what follows from this.

Review: Matrix Multiplication

First of all, let's agree on notation. A matrix $A$ of size $\left[ m\times n \right]$ is simply a table of numbers with exactly $m$ rows and $n$ columns:

\=\underbrace(\left[ \begin(matrix) ((a)_(11)) & ((a)_(12)) & ... & ((a)_(1n)) \\ (( a)_(21)) & ((a)_(22)) & ... & ((a)_(2n)) \\ ... & ... & ... & ... \\ ((a)_(m1)) & ((a)_(m2)) & ... & ((a)_(mn)) \\\end(matrix) \right])_(n)\]

To avoid accidentally mixing up rows and columns (believe me, in an exam you can confuse a one with a two, let alone some rows), just look at the picture:

Determining indices for matrix cells

What's happening? If you place the standard coordinate system $OXY$ in the upper left corner and direct the axes so that they cover the entire matrix, then each cell of this matrix can be uniquely associated with coordinates $\left(x;y \right)$ - this will be the row number and column number.

Why is the coordinate system placed in the upper left corner? Yes, because it is from there that we begin to read any texts. It's very easy to remember.

Why is the $x$ axis directed downwards and not to the right? Again, it's simple: take a standard coordinate system (the $x$ axis goes to the right, the $y$ axis goes up) and rotate it so that it covers the matrix. This is a 90 degree clockwise rotation - we see the result in the picture.

In general, we have figured out how to determine the indices of matrix elements. Now let's look at multiplication.

Definition. Matrices $A=\left[ m\times n \right]$ and $B=\left[ n\times k \right]$, when the number of columns in the first coincides with the number of rows in the second, are called consistent.

Exactly in that order. One can be confused and say that the matrices $A$ and $B$ form an ordered pair $\left(A;B \right)$: if they are consistent in this order, then it is not at all necessary that $B$ and $A$ those. the pair $\left(B;A \right)$ is also consistent.

Only matched matrices can be multiplied.

Definition. The product of matched matrices $A=\left[ m\times n \right]$ and $B=\left[ n\times k \right]$ is the new matrix $C=\left[ m\times k \right]$ , the elements of which $((c)_(ij))$ are calculated according to the formula:

\[((c)_(ij))=\sum\limits_(k=1)^(n)(((a)_(ik)))\cdot ((b)_(kj))\]

In other words: to get the element $((c)_(ij))$ of the matrix $C=A\cdot B$, you need to take the $i$-row of the first matrix, the $j$-th column of the second matrix, and then multiply in pairs elements from this row and column. Add up the results.

Yes, that’s such a harsh definition. Several facts immediately follow from it:

1. Matrix multiplication, generally speaking, is non-commutative: $A\cdot B\ne B\cdot A$;
2. However, multiplication is associative: $\left(A\cdot B \right)\cdot C=A\cdot \left(B\cdot C \right)$;
3. And even distributively: $\left(A+B \right)\cdot C=A\cdot C+B\cdot C$;
4. And once again distributively: $A\cdot \left(B+C \right)=A\cdot B+A\cdot C$.

The distributivity of multiplication had to be described separately for the left and right sum factor precisely because of the non-commutativity of the multiplication operation.

If it turns out that $A\cdot B=B\cdot A$, such matrices are called commutative.

Among all the matrices that are multiplied by something there, there are special ones - those that, when multiplied by any matrix $A$, again give $A$:

Definition. A matrix $E$ is called identity if $A\cdot E=A$ or $E\cdot A=A$. In the case of a square matrix $A$ we can write:

The identity matrix is ​​a frequent guest in solving matrix equations. And in general, a frequent guest in the world of matrices. :)

And because of this $E$, someone came up with all the nonsense that will be written next.

What is an inverse matrix

Since matrix multiplication is a very labor-intensive operation (you have to multiply a bunch of rows and columns), the concept of an inverse matrix also turns out to be not the most trivial. And requiring some explanation.

Key Definition

Well, it's time to know the truth.

Definition. A matrix $B$ is called the inverse of a matrix $A$ if

The inverse matrix is ​​denoted by $((A)^(-1))$ (not to be confused with the degree!), so the definition can be rewritten as follows:

It would seem that everything is extremely simple and clear. But when analyzing this definition, several questions immediately arise:

1. Does an inverse matrix always exist? And if not always, then how to determine: when it exists and when it does not?
2. And who said that there is exactly one such matrix? What if for some initial matrix $A$ there is a whole crowd of inverses?
3. What do all these “reverses” look like? And how, exactly, should we count them?

As for calculation algorithms, we will talk about this a little later. But we will answer the remaining questions right now. Let us formulate them in the form of separate statements-lemmas.

Basic properties

Let's start with how the matrix $A$ should, in principle, look in order for $((A)^(-1))$ to exist for it. Now we will make sure that both of these matrices must be square and of the same size: $\left[ n\times n \right]$.

Lemma 1. Given a matrix $A$ and its inverse $((A)^(-1))$. Then both of these matrices are square, and of the same order $n$.

Proof. It's simple. Let the matrix $A=\left[ m\times n \right]$, $((A)^(-1))=\left[ a\times b \right]$. Since the product $A\cdot ((A)^(-1))=E$ exists by definition, the matrices $A$ and $((A)^(-1))$ are consistent in the order shown:

\[\begin(align) & \left[ m\times n \right]\cdot \left[ a\times b \right]=\left[ m\times b \right] \\ & n=a \end( align)\]

This is a direct consequence of the matrix multiplication algorithm: the coefficients $n$ and $a$ are “transit” and must be equal.

At the same time, the inverse multiplication is also defined: $((A)^(-1))\cdot A=E$, therefore the matrices $((A)^(-1))$ and $A$ are also consistent in the specified order:

\[\begin(align) & \left[ a\times b \right]\cdot \left[ m\times n \right]=\left[ a\times n \right] \\ & b=m \end( align)\]

Thus, without loss of generality, we can assume that $A=\left[ m\times n \right]$, $((A)^(-1))=\left[ n\times m \right]$. However, according to the definition of $A\cdot ((A)^(-1))=((A)^(-1))\cdot A$, therefore the sizes of the matrices strictly coincide:

\[\begin(align) & \left[ m\times n \right]=\left[ n\times m \right] \\ & m=n \end(align)\]

So it turns out that all three matrices - $A$, $((A)^(-1))$ and $E$ - are square matrices of size $\left[ n\times n \right]$. The lemma is proven.

Well, that's already good. We see that only square matrices are invertible. Now let's make sure that the inverse matrix is ​​always the same.

Lemma 2. Given a matrix $A$ and its inverse $((A)^(-1))$. Then this inverse matrix is ​​the only one.

Proof. Let's go by contradiction: let the matrix $A$ have at least two inverses - $B$ and $C$. Then, according to definition, the following equalities are true:

\[\begin(align) & A\cdot B=B\cdot A=E; \\ & A\cdot C=C\cdot A=E. \\ \end(align)\]

From Lemma 1 we conclude that all four matrices - $A$, $B$, $C$ and $E$ - are squares of the same order: $\left[ n\times n \right]$. Therefore, the product is defined:

Since matrix multiplication is associative (but not commutative!), we can write:

\[\begin(align) & B\cdot A\cdot C=\left(B\cdot A \right)\cdot C=E\cdot C=C; \\ & B\cdot A\cdot C=B\cdot \left(A\cdot C \right)=B\cdot E=B; \\ & B\cdot A\cdot C=C=B\Rightarrow B=C. \\ \end(align)\]

We received only possible variant: two instances of the inverse matrix are equal. The lemma is proven.

The above arguments repeat almost verbatim the proof of the uniqueness of the inverse element for all real numbers$b\ne 0$. The only significant addition is taking into account the dimension of matrices.

However, we still do not know anything about whether every square matrix is ​​invertible. Here the determinant comes to our aid - this is a key characteristic for all square matrices.

Lemma 3. Given a matrix $A$. If its inverse matrix $((A)^(-1))$ exists, then the determinant of the original matrix is ​​nonzero:

\[\left| A\right|\ne 0\]

Proof. We already know that $A$ and $((A)^(-1))$ are square matrices of size $\left[ n\times n \right]$. Therefore, for each of them we can calculate the determinant: $\left| A\right|$ and $\left| ((A)^(-1)) \right|$. However, the determinant of a product is equal to the product of the determinants:

\[\left| A\cdot B \right|=\left| A \right|\cdot \left| B \right|\Rightarrow \left| A\cdot ((A)^(-1)) \right|=\left| A \right|\cdot \left| ((A)^(-1)) \right|\]

But according to the definition, $A\cdot ((A)^(-1))=E$, and the determinant of $E$ is always equal to 1, so

\[\begin(align) & A\cdot ((A)^(-1))=E; \\ & \left| A\cdot ((A)^(-1)) \right|=\left| E\right|; \\ & \left| A \right|\cdot \left| ((A)^(-1)) \right|=1. \\ \end(align)\]

The product of two numbers is equal to one only if each of these numbers is non-zero:

\[\left| A \right|\ne 0;\quad \left| ((A)^(-1)) \right|\ne 0.\]

So it turns out that $\left| A \right|\ne 0$. The lemma is proven.

In fact, this requirement is quite logical. Now we will analyze the algorithm for finding the inverse matrix - and it will become completely clear why, with a zero determinant, no inverse matrix in principle can exist.

But first, let’s formulate an “auxiliary” definition:

Definition. A singular matrix is ​​a square matrix of size $\left[ n\times n \right]$ whose determinant is zero.

Thus, we can claim that every invertible matrix is ​​non-singular.

How to find the inverse of a matrix

Now we will consider a universal algorithm for finding inverse matrices. In general, there are two generally accepted algorithms, and we will also consider the second one today.

The one that will be discussed now is very effective for matrices of size $\left[ 2\times 2 \right]$ and - partially - size $\left[ 3\times 3 \right]$. But starting from the size $\left[ 4\times 4 \right]$ it is better not to use it. Why - now you will understand everything yourself.

Algebraic additions

Get ready. Now there will be pain. No, don’t worry: a beautiful nurse in a skirt, stockings with lace will not come to you and give you an injection in the buttock. Everything is much more prosaic: they are coming to you algebraic additions and Her Majesty "Union Matrix".

Let's start with the main thing. Let there be a square matrix of size $A=\left[ n\times n \right]$, whose elements are called $((a)_(ij))$. Then for each such element we can define an algebraic complement:

Definition. Algebraic complement $((A)_(ij))$ to the element $((a)_(ij))$ located in the $i$th row and $j$th column of the matrix $A=\left[ n \times n \right]$ is a construction of the form

\[((A)_(ij))=((\left(-1 \right))^(i+j))\cdot M_(ij)^(*)\]

Where $M_(ij)^(*)$ is the determinant of the matrix obtained from the original $A$ by deleting the same $i$th row and $j$th column.

Again. The algebraic complement to a matrix element with coordinates $\left(i;j \right)$ is denoted as $((A)_(ij))$ and is calculated according to the scheme:

1. First, we delete the $i$-row and $j$-th column from the original matrix. We obtain a new square matrix, and we denote its determinant as $M_(ij)^(*)$.
2. Then we multiply this determinant by $((\left(-1 \right))^(i+j))$ - at first this expression may seem mind-blowing, but in essence we are simply figuring out the sign in front of $M_(ij)^(*) $.
3. We count and get a specific number. Those. the algebraic addition is precisely a number, and not some new matrix, etc.

The matrix $M_(ij)^(*)$ itself is called an additional minor to the element $((a)_(ij))$. And in this sense, the above definition of an algebraic complement is a special case of a more complex definition - what we looked at in the lesson about the determinant.

Important note. Actually, in “adult” mathematics, algebraic additions are defined as follows:

1. We take $k$ rows and $k$ columns in a square matrix. At their intersection we get a matrix of size $\left[ k\times k \right]$ - its determinant is called a minor of order $k$ and is denoted $((M)_(k))$.
2. Then we cross out these “selected” $k$ rows and $k$ columns. Once again you get a square matrix - its determinant is called an additional minor and is denoted $M_(k)^(*)$.
3. Multiply $M_(k)^(*)$ by $((\left(-1 \right))^(t))$, where $t$ is (attention now!) the sum of the numbers of all selected rows and columns . This will be the algebraic addition.

Look at the third step: there is actually a sum of $2k$ terms! Another thing is that for $k=1$ we will get only 2 terms - these will be the same $i+j$ - the “coordinates” of the element $((a)_(ij))$ for which we are looking for an algebraic complement.

So today we're using a slightly simplified definition. But as we will see later, it will be more than enough. The following thing is much more important:

Definition. The allied matrix $S$ to the square matrix $A=\left[ n\times n \right]$ is a new matrix of size $\left[ n\times n \right]$, which is obtained from $A$ by replacing $(( a)_(ij))$ by algebraic additions $((A)_(ij))$:

\\Rightarrow S=\left[ \begin(matrix) ((A)_(11)) & ((A)_(12)) & ... & ((A)_(1n)) \\ (( A)_(21)) & ((A)_(22)) & ... & ((A)_(2n)) \\ ... & ... & ... & ... \\ ((A)_(n1)) & ((A)_(n2)) & ... & ((A)_(nn)) \\\end(matrix) \right]\]

The first thought that arises at the moment of realizing this definition is “how much will have to be counted!” Relax: you will have to count, but not that much. :)

Well, all this is very nice, but why is it necessary? But why.

Main theorem

Let's go back a little. Remember, in Lemma 3 it was stated that the invertible matrix $A$ is always non-singular (that is, its determinant is non-zero: $\left| A \right|\ne 0$).

So, the opposite is also true: if the matrix $A$ is not singular, then it is always invertible. And there is even a search scheme for $((A)^(-1))$. Check it out:

Inverse matrix theorem. Let a square matrix $A=\left[ n\times n \right]$ be given, and its determinant is nonzero: $\left| A \right|\ne 0$. Then the inverse matrix $((A)^(-1))$ exists and is calculated by the formula:

\[((A)^(-1))=\frac(1)(\left| A \right|)\cdot ((S)^(T))\]

And now - everything is the same, but in legible handwriting. To find the inverse matrix, you need:

1. Calculate the determinant $\left| A \right|$ and make sure it is non-zero.
2. Construct the union matrix $S$, i.e. count 100500 algebraic additions $((A)_(ij))$ and place them in place $((a)_(ij))$.
3. Transpose this matrix $S$, and then multiply it by some number $q=(1)/(\left| A \right|)\;$.

That's all! The inverse matrix $((A)^(-1))$ has been found. Let's look at examples:

\[\left[ \begin(matrix) 3 & 1 \\ 5 & 2 \\\end(matrix) \right]\]

Solution. Let's check the reversibility. Let's calculate the determinant:

\[\left| A\right|=\left| \begin(matrix) 3 & 1 \\ 5 & 2 \\\end(matrix) \right|=3\cdot 2-1\cdot 5=6-5=1\]

The determinant is different from zero. This means the matrix is ​​invertible. Let's create a union matrix:

Let's calculate the algebraic additions:

\[\begin(align) & ((A)_(11))=((\left(-1 \right))^(1+1))\cdot \left| 2 \right|=2; \\ & ((A)_(12))=((\left(-1 \right))^(1+2))\cdot \left| 5 \right|=-5; \\ & ((A)_(21))=((\left(-1 \right))^(2+1))\cdot \left| 1 \right|=-1; \\ & ((A)_(22))=((\left(-1 \right))^(2+2))\cdot \left| 3\right|=3. \\ \end(align)\]

Please note: the determinants |2|, |5|, |1| and |3| are determinants of matrices of size $\left[ 1\times 1 \right]$, and not modules. Those. If there were negative numbers in the determinants, there is no need to remove the “minus”.

In total, our union matrix looks like this:

\[((A)^(-1))=\frac(1)(\left| A \right|)\cdot ((S)^(T))=\frac(1)(1)\cdot ( (\left[ \begin(array)(*(35)(r)) 2 & -5 \\ -1 & 3 \\\end(array) \right])^(T))=\left[ \begin (array)(*(35)(r)) 2 & -1 \\ -5 & 3 \\\end(array) \right]\]

OK it's all over Now. The problem is solved.

Answer. $\left[ \begin(array)(*(35)(r)) 2 & -1 \\ -5 & 3 \\\end(array) \right]$

Task. Find the inverse matrix:

\[\left[ \begin(array)(*(35)(r)) 1 & -1 & 2 \\ 0 & 2 & -1 \\ 1 & 0 & 1 \\\end(array) \right] \]

Solution. We calculate the determinant again:

\[\begin(align) & \left| \begin(array)(*(35)(r)) 1 & -1 & 2 \\ 0 & 2 & -1 \\ 1 & 0 & 1 \\\end(array) \right|=\begin(matrix ) \left(1\cdot 2\cdot 1+\left(-1 \right)\cdot \left(-1 \right)\cdot 1+2\cdot 0\cdot 0 \right)- \\ -\left (2\cdot 2\cdot 1+\left(-1 \right)\cdot 0\cdot 1+1\cdot \left(-1 \right)\cdot 0 \right) \\\end(matrix)= \ \ & =\left(2+1+0 \right)-\left(4+0+0 \right)=-1\ne 0. \\ \end(align)\]

The determinant is nonzero—the matrix is ​​invertible. But now it’s going to be really tough: we need to count as many as 9 (nine, motherfucker!) algebraic additions. And each of them will contain the determinant $\left[ 2\times 2 \right]$. Flew:

\[\begin(matrix) ((A)_(11))=((\left(-1 \right))^(1+1))\cdot \left| \begin(matrix) 2 & -1 \\ 0 & 1 \\\end(matrix) \right|=2; \\ ((A)_(12))=((\left(-1 \right))^(1+2))\cdot \left| \begin(matrix) 0 & -1 \\ 1 & 1 \\\end(matrix) \right|=-1; \\ ((A)_(13))=((\left(-1 \right))^(1+3))\cdot \left| \begin(matrix) 0 & 2 \\ 1 & 0 \\\end(matrix) \right|=-2; \\ ... \\ ((A)_(33))=((\left(-1 \right))^(3+3))\cdot \left| \begin(matrix) 1 & -1 \\ 0 & 2 \\\end(matrix) \right|=2; \\ \end(matrix)\]

In short, the union matrix will look like this:

Therefore, the inverse matrix will be:

\[((A)^(-1))=\frac(1)(-1)\cdot \left[ \begin(matrix) 2 & -1 & -2 \\ 1 & -1 & -1 \\ -3 & 1 & 2 \\\end(matrix) \right]=\left[ \begin(array)(*(35)(r))-2 & -1 & 3 \\ 1 & 1 & -1 \ \2 & 1 & -2 \\\end(array) \right]\]

That's it. Here is the answer.

Answer. $\left[ \begin(array)(*(35)(r)) -2 & -1 & 3 \\ 1 & 1 & -1 \\ 2 & 1 & -2 \\\end(array) \right ]$

As you can see, at the end of each example we performed a check. In this regard, an important note:

Don't be lazy to check. Multiply the original matrix by the found inverse matrix - you should get $E$.

Performing this check is much easier and faster than looking for an error in further calculations when, for example, you are solving a matrix equation.

Alternative way

As I said, the inverse matrix theorem works great for sizes $\left[ 2\times 2 \right]$ and $\left[ 3\times 3 \right]$ (in the latter case, it’s not so “great” "), but for matrices large sizes the sadness begins.

But don’t worry: there is an alternative algorithm with which you can calmly find the inverse even for the matrix $\left[ 10\times 10 \right]$. But, as often happens, to consider this algorithm we need a little theoretical background.

Elementary transformations

Among all possible matrix transformations, there are several special ones - they are called elementary. There are exactly three such transformations:

1. Multiplication. You can take the $i$th row (column) and multiply it by any number $k\ne 0$;
2. Addition. Add to the $i$-th row (column) any other $j$-th row (column), multiplied by any number $k\ne 0$ (you can, of course, do $k=0$, but what's the point? ? Nothing will change).
3. Rearrangement. Take the $i$th and $j$th rows (columns) and swap places.

Why these transformations are called elementary (for large matrices they do not look so elementary) and why there are only three of them - these questions are beyond the scope of today's lesson. Therefore, we will not go into details.

Another thing is important: we have to perform all these perversions on the adjoint matrix. Yes, yes: you heard right. Now there will be one more definition - the last one in today's lesson.

Adjoint matrix

Surely at school you solved systems of equations using the addition method. Well, there, subtract another from one line, multiply some line by a number - that’s all.

So: now everything will be the same, but in an “adult” way. Ready?

Definition. Let a matrix $A=\left[ n\times n \right]$ and an identity matrix $E$ of the same size $n$ be given. Then the adjoint matrix $\left[ A\left| E\right. \right]$ is a new matrix of size $\left[ n\times 2n \right]$ that looks like this:

\[\left[ A\left| E\right. \right]=\left[ \begin(array)(rrrr|rrrr)((a)_(11)) & ((a)_(12)) & ... & ((a)_(1n)) & 1 & 0 & ... & 0 \\((a)_(21)) & ((a)_(22)) & ... & ((a)_(2n)) & 0 & 1 & ... & 0 \\... & ... & ... & ... & ... & ... & ... & ... \\((a)_(n1)) & ((a)_(n2)) & ... & ((a)_(nn)) & 0 & 0 & ... & 1 \\\end(array) \right]\]

In short, we take the matrix $A$, and on the right we assign to it the identity matrix $E$ the right size, we separate them with a vertical line for beauty - here you have the attached one. :)

What's the catch? Here's what:

Theorem. Let the matrix $A$ be invertible. Consider the adjoint matrix $\left[ A\left| E\right. \right]$. If using elementary string conversions bring it to the form $\left[ E\left| B\right. \right]$, i.e. by multiplying, subtracting and rearranging rows to obtain from $A$ the matrix $E$ on the right, then the matrix $B$ obtained on the left is the inverse of $A$:

\[\left[ A\left| E\right. \right]\to \left[ E\left| B\right. \right]\Rightarrow B=((A)^(-1))\]

It's that simple! In short, the algorithm for finding the inverse matrix looks like this:

1. Write the adjoint matrix $\left[ A\left| E\right. \right]$;
2. Perform elementary string conversions until $E$ appears instead of $A$;
3. Of course, something will also appear on the left - a certain matrix $B$. This will be the opposite;
4. PROFIT!:)

Of course, this is much easier said than done. So let's look at a couple of examples: for sizes $\left[ 3\times 3 \right]$ and $\left[ 4\times 4 \right]$.

Task. Find the inverse matrix:

\[\left[ \begin(array)(*(35)(r)) 1 & 5 & 1 \\ 3 & 2 & 1 \\ 6 & -2 & 1 \\\end(array) \right]\ ]

Solution. We create the adjoint matrix:

\[\left[ \begin(array)(rrr|rrr) 1 & 5 & 1 & 1 & 0 & 0 \\ 3 & 2 & 1 & 0 & 1 & 0 \\ 6 & -2 & 1 & 0 & 0 & 1 \\\end(array) \right]\]

Since the last column of the original matrix is ​​filled with ones, subtract the first row from the rest:

\[\begin(align) & \left[ \begin(array)(rrr|rrr) 1 & 5 & 1 & 1 & 0 & 0 \\ 3 & 2 & 1 & 0 & 1 & 0 \\ 6 & - 2 & 1 & 0 & 0 & 1 \\\end(array) \right]\begin(matrix) \downarrow \\ -1 \\ -1 \\\end(matrix)\to \\ & \to \left [ \begin(array)(rrr|rrr) 1 & 5 & 1 & 1 & 0 & 0 \\ 2 & -3 & 0 & -1 & 1 & 0 \\ 5 & -7 & 0 & -1 & 0 & 1 \\\end(array) \right] \\ \end(align)\]

There are no more units, except for the first line. But we don’t touch it, otherwise the newly removed units will begin to “multiply” in the third column.

But we can subtract the second line twice from the last - we get one in the lower left corner:

\[\begin(align) & \left[ \begin(array)(rrr|rrr) 1 & 5 & 1 & 1 & 0 & 0 \\ 2 & -3 & 0 & -1 & 1 & 0 \\ 5 & -7 & 0 & -1 & 0 & 1 \\\end(array) \right]\begin(matrix) \ \\ \downarrow \\ -2 \\\end(matrix)\to \\ & \left [ \begin(array)(rrr|rrr) 1 & 5 & 1 & 1 & 0 & 0 \\ 2 & -3 & 0 & -1 & 1 & 0 \\ 1 & -1 & 0 & 1 & -2 & 1 \\\end(array) \right] \\ \end(align)\]

Now we can subtract the last row from the first and twice from the second - this way we “zero” the first column:

\[\begin(align) & \left[ \begin(array)(rrr|rrr) 1 & 5 & 1 & 1 & 0 & 0 \\ 2 & -3 & 0 & -1 & 1 & 0 \\ 1 & -1 & 0 & 1 & -2 & 1 \\\end(array) \right]\begin(matrix) -1 \\ -2 \\ \uparrow \\\end(matrix)\to \\ & \ to \left[ \begin(array)(rrr|rrr) 0 & 6 & 1 & 0 & 2 & -1 \\ 0 & -1 & 0 & -3 & 5 & -2 \\ 1 & -1 & 0 & 1 & -2 & 1 \\\end(array) \right] \\ \end(align)\]

Multiply the second line by −1, and then subtract it 6 times from the first and add 1 time to the last:

\[\begin(align) & \left[ \begin(array)(rrr|rrr) 0 & 6 & 1 & 0 & 2 & -1 \\ 0 & -1 & 0 & -3 & 5 & -2 \ \ 1 & -1 & 0 & 1 & -2 & 1 \\\end(array) \right]\begin(matrix) \ \\ \left| \cdot \left(-1 \right) \right. \\ \ \\\end(matrix)\to \\ & \to \left[ \begin(array)(rrr|rrr) 0 & 6 & 1 & 0 & 2 & -1 \\ 0 & 1 & 0 & 3 & -5 & 2 \\ 1 & -1 & 0 & 1 & -2 & 1 \\\end(array) \right]\begin(matrix) -6 \\ \updownarrow \\ +1 \\\end (matrix)\to \\ & \to \left[ \begin(array)(rrr|rrr) 0 & 0 & 1 & -18 & 32 & -13 \\ 0 & 1 & 0 & 3 & -5 & 2 \\ 1 & 0 & 0 & 4 & -7 & 3 \\\end(array) \right] \\ \end(align)\]

All that remains is to swap lines 1 and 3:

\[\left[ \begin(array)(rrr|rrr) 1 & 0 & 0 & 4 & -7 & 3 \\ 0 & 1 & 0 & 3 & -5 & 2 \\ 0 & 0 & 1 & - 18 & 32 & -13 \\\end(array) \right]\]

Ready! On the right is the required inverse matrix.

Answer. $\left[ \begin(array)(*(35)(r))4 & -7 & 3 \\ 3 & -5 & 2 \\ -18 & 32 & -13 \\\end(array) \right ]$

Task. Find the inverse matrix:

\[\left[ \begin(matrix) 1 & 4 & 2 & 3 \\ 1 & -2 & 1 & -2 \\ 1 & -1 & 1 & 1 \\ 0 & -10 & -2 & -5 \\\end(matrix) \right]\]

Solution. We compose the adjoint again:

\[\left[ \begin(array)(rrrr|rrrr) 1 & 4 & 2 & 3 & 1 & 0 & 0 & 0 \\ 1 & -2 & 1 & -2 & 0 & 1 & 0 & 0 \ \ 1 & -1 & 1 & 1 & 0 & 0 & 1 & 0 \\ 0 & -10 & -2 & -5 & 0 & 0 & 0 & 1 \\\end(array) \right]\]

Let's cry a little, be sad about how much we have to count now... and start counting. First, let’s “zero out” the first column by subtracting row 1 from rows 2 and 3:

\[\begin(align) & \left[ \begin(array)(rrrr|rrrr) 1 & 4 & 2 & 3 & 1 & 0 & 0 & 0 \\ 1 & -2 & 1 & -2 & 0 & 1 & 0 & 0 \\ 1 & -1 & 1 & 1 & 0 & 0 & 1 & 0 \\ 0 & -10 & -2 & -5 & 0 & 0 & 0 & 1 \\\end(array) \right]\begin(matrix) \downarrow \\ -1 \\ -1 \\ \ \\\end(matrix)\to \\ & \to \left[ \begin(array)(rrrr|rrrr) 1 & 4 & 2 & 3 & 1 & 0 & 0 & 0 \\ 0 & -6 & -1 & -5 & -1 & 1 & 0 & 0 \\ 0 & -5 & -1 & -2 & -1 & 0 & 1 & 0 \\ 0 & -10 & -2 & -5 & 0 & 0 & 0 & 1 \\\end(array) \right] \\ \end(align)\]

We see too many “cons” in lines 2-4. Multiply all three rows by −1, and then burn out the third column by subtracting row 3 from the rest:

\[\begin(align) & \left[ \begin(array)(rrrr|rrrr) 1 & 4 & 2 & 3 & 1 & 0 & 0 & 0 \\ 0 & -6 & -1 & -5 & - 1 & 1 & 0 & 0 \\ 0 & -5 & -1 & -2 & -1 & 0 & 1 & 0 \\ 0 & -10 & -2 & -5 & 0 & 0 & 0 & 1 \\ \end(array) \right]\begin(matrix) \ \\ \left| \cdot \left(-1 \right) \right. \\ \left| \cdot \left(-1 \right) \right. \\ \left| \cdot \left(-1 \right) \right. \\\end(matrix)\to \\ & \to \left[ \begin(array)(rrrr|rrrr) 1 & 4 & 2 & 3 & 1 & 0 & 0 & 0 \\ 0 & 6 & 1 & 5 & ​​1 & -1 & 0 & 0 \\ 0 & 5 & 1 & 2 & 1 & 0 & -1 & 0 \\ 0 & 10 & 2 & 5 & 0 & 0 & 0 & -1 \\\end (array) \right]\begin(matrix) -2 \\ -1 \\ \updownarrow \\ -2 \\\end(matrix)\to \\ & \to \left[ \begin(array)(rrrr| rrrr) 1 & -6 & 0 & -1 & -1 & 0 & 2 & 0 \\ 0 & 1 & 0 & 3 & 0 & -1 & 1 & 0 \\ 0 & 5 & 1 & 2 & 1 & 0 & -1 & 0 \\ 0 & 0 & 0 & 1 & -2 & 0 & 2 & -1 \\\end(array) \right] \\ \end(align)\]

Now is the time to “fry” the last column of the original matrix: subtract row 4 from the rest:

\[\begin(align) & \left[ \begin(array)(rrrr|rrrr) 1 & -6 & 0 & -1 & -1 & 0 & 2 & 0 \\ 0 & 1 & 0 & 3 & 0 & -1 & 1 & 0 \\ 0 & 5 & 1 & 2 & 1 & 0 & -1 & 0 \\ 0 & 0 & 0 & 1 & -2 & 0 & 2 & -1 \\\end(array ) \right]\begin(matrix) +1 \\ -3 \\ -2 \\ \uparrow \\\end(matrix)\to \\ & \to \left[ \begin(array)(rrrr|rrrr) 1 & -6 & 0 & 0 & -3 & 0 & 4 & -1 \\ 0 & 1 & 0 & 0 & 6 & -1 & -5 & 3 \\ 0 & 5 & 1 & 0 & 5 & 0 & -5 & 2 \\ 0 & 0 & 0 & 1 & -2 & 0 & 2 & -1 \\\end(array) \right] \\ \end(align)\]

Final throw: “burn out” the second column by subtracting line 2 from lines 1 and 3:

\[\begin(align) & \left[ \begin(array)(rrrr|rrrr) 1 & -6 & 0 & 0 & -3 & 0 & 4 & -1 \\ 0 & 1 & 0 & 0 & 6 & -1 & -5 & 3 \\ 0 & 5 & 1 & 0 & 5 & 0 & -5 & 2 \\ 0 & 0 & 0 & 1 & -2 & 0 & 2 & -1 \\\end( array) \right]\begin(matrix) 6 \\ \updownarrow \\ -5 \\ \ \\\end(matrix)\to \\ & \to \left[ \begin(array)(rrrr|rrrr) 1 & 0 & 0 & 0 & 33 & -6 & -26 & -17 \\ 0 & 1 & 0 & 0 & 6 & -1 & -5 & 3 \\ 0 & 0 & 1 & 0 & -25 & 5 & 20 & -13 \\ 0 & 0 & 0 & 1 & -2 & 0 & 2 & -1 \\\end(array) \right] \\ \end(align)\]

And again the identity matrix is ​​on the left, which means the inverse is on the right. :)

Answer. $\left[ \begin(matrix) 33 & -6 & -26 & 17 \\ 6 & -1 & -5 & 3 \\ -25 & 5 & 20 & -13 \\ -2 & 0 & 2 & - 1 \\\end(matrix) \right]$

Matrix A -1 is called the inverse matrix with respect to matrix A if A*A -1 = E, where E is the identity matrix of the nth order. An inverse matrix can only exist for square matrices.

Purpose of the service. Using this service online you can find algebraic complements, transposed matrix A T, allied matrix and inverse matrix. The decision is carried out directly on the website (online) and is free. The calculation results are presented in a report Word format and in Excel format (i.e. it is possible to check the solution). see design example.

Instructions. To obtain a solution, it is necessary to specify the dimension of the matrix. Next, fill out matrix A in the new dialog box.

Matrix dimension 2 3 4 5 6 7 8 9 10

See also Inverse matrix using the Jordano-Gauss method

Algorithm for finding the inverse matrix

1. Finding the transposed matrix A T .
2. Definition of algebraic complements. Replace each element of the matrix with its algebraic complement.
3. Compiling an inverse matrix from algebraic additions: each element of the resulting matrix is ​​divided by the determinant of the original matrix. The resulting matrix is ​​the inverse of the original matrix.

Next algorithm for finding the inverse matrix similar to the previous one except for some steps: first the algebraic complements are calculated, and then the allied matrix C is determined.

1. Determine whether the matrix is ​​square. If not, then there is no inverse matrix for it.
2. Calculation of the determinant of the matrix A. If it is not equal to zero, we continue the solution, otherwise the inverse matrix does not exist.
3. Definition of algebraic complements.
4. Filling out the union (mutual, adjoint) matrix C .
5. Compiling an inverse matrix from algebraic additions: each element of the adjoint matrix C is divided by the determinant of the original matrix. The resulting matrix is ​​the inverse of the original matrix.
6. They do a check: they multiply the original and the resulting matrices. The result should be an identity matrix.

Example No. 1. Let's write the matrix in the form:

Algebraic additions.

A 1,1 = (-1) 1+1

-1 -2 5 4

∆ 1,1 = (-1 4-5 (-2)) = 6

A 1,2 = (-1) 1+2

2 -2 -2 4

∆ 1,2 = -(2 4-(-2 (-2))) = -4

A 1.3 = (-1) 1+3

2 -1 -2 5

∆ 1,3 = (2 5-(-2 (-1))) = 8

A 2,1 = (-1) 2+1

2 3 5 4

∆ 2,1 = -(2 4-5 3) = 7

A 2,2 = (-1) 2+2

-1 3 -2 4

∆ 2,2 = (-1 4-(-2 3)) = 2

A 2,3 = (-1) 2+3

-1 2 -2 5

∆ 2,3 = -(-1 5-(-2 2)) = 1

A 3.1 = (-1) 3+1

2 3 -1 -2

∆ 3,1 = (2 (-2)-(-1 3)) = -1

A 3.2 = (-1) 3+2

-1 3 2 -2

∆ 3,2 = -(-1 (-2)-2 3) = 4

A 3.3 = (-1) 3+3

-1 2 2 -1

∆ 3,3 = (-1 (-1)-2 2) = -3
Then inverse matrix can be written as:

A -1 = 1/10

6 -4 8 7 2 1 -1 4 -3

A -1 =

0,6 -0,4 0,8 0,7 0,2 0,1 -0,1 0,4 -0,3

Another algorithm for finding the inverse matrix

Let us present another scheme for finding the inverse matrix.

1. Find the determinant of a given square matrix A.
2. We find algebraic complements to all elements of the matrix A.
3. We write algebraic additions of row elements to columns (transposition).
4. We divide each element of the resulting matrix by the determinant of the matrix A.

As we can see, the transposition operation can be applied both at the beginning, on the original matrix, and at the end, on the resulting algebraic additions.

A special case: The inverse of the identity matrix E is the identity matrix E.

Finding the inverse matrix.

In this article we will understand the concept of an inverse matrix, its properties and methods of finding. Let us dwell in detail on solving examples in which it is necessary to construct an inverse matrix for a given one.

Page navigation.

Inverse matrix - definition.

Finding the inverse matrix using a matrix from algebraic complements.

Properties of an inverse matrix.

Finding the inverse matrix using the Gauss-Jordan method.

Finding the elements of the inverse matrix by solving the corresponding systems of linear algebraic equations.

Inverse matrix - definition.

The concept of an inverse matrix is ​​introduced only for square matrices whose determinant is nonzero, that is, for non-singular square matrices.

Definition.

Matrixcalled the inverse of a matrix, whose determinant is different from zero if the equalities are true , Where E– unit order matrix n on n.

Finding the inverse matrix using a matrix from algebraic complements.

How to find the inverse matrix for a given one?

First, we need the concepts transposed matrix, matrix minor and algebraic complement of a matrix element.

Definition.

Minorkth order matrices A order m on n is the determinant of the order matrix k on k, which is obtained from the matrix elements A located in the selected k lines and k columns. ( k does not exceed the smallest number m or n).

Minor (n-1)th order, which is composed of elements of all rows except i-th, and all columns except jth, square matrix A order n on n let's denote it as .

In other words, the minor is obtained from a square matrix A order n on n by crossing out elements i-th lines and jth column.

For example, let's write, minor 2nd order, which is obtained from the matrix selecting elements of its second, third rows and first, third columns . We will also show the minor, which is obtained from the matrix by crossing out the second line and third column . Let us illustrate the construction of these minors: and .

Definition.

Algebraic complement element of a square matrix is ​​called minor (n-1)th order, which is obtained from the matrix A, crossing out elements of it i-th lines and jth column multiplied by .

The algebraic complement of an element is denoted as . Thus, .

For example, for the matrix the algebraic complement of an element is .

Secondly, we will need two properties of the determinant, which we discussed in the section calculating the determinant of a matrix:

Based on these properties of the determinant, the definition operations of multiplying a matrix by a number and the concept of an inverse matrix is ​​true: , where is a transposed matrix whose elements are algebraic complements.

Matrix is indeed the inverse of the matrix A, since the equalities are satisfied . Let's show it

Let's compose algorithm for finding the inverse matrix using equality .

Let's look at the algorithm for finding the inverse matrix using an example.

Example.

Given a matrix . Find the inverse matrix.

Solution.

Let's calculate the determinant of the matrix A, decomposing it into the elements of the third column:

The determinant is nonzero, so the matrix A reversible.

Let's find a matrix of algebraic additions:

That's why

Let's transpose the matrix from algebraic additions:

Now we find the inverse matrix as :

Let's check the result:

Equalities are satisfied, therefore, the inverse matrix is ​​found correctly.

Properties of an inverse matrix.

The concept of an inverse matrix, equality , definitions of operations on matrices and properties of the determinant of a matrix make it possible to justify the following properties of inverse matrix:

Finding the elements of the inverse matrix by solving the corresponding systems of linear algebraic equations.

Let's consider another way to find the inverse matrix for a square matrix A order n on n.

This method is based on the solution n systems of linear inhomogeneous algebraic equations with n unknown. The unknown variables in these systems of equations are the elements of the inverse matrix.

The idea is very simple. Let us denote the inverse matrix as X, that is, . Since by definition of the inverse matrix, then

Equating the corresponding elements by columns, we get n systems of linear equations

We solve them in any way and form an inverse matrix from the found values.

Let's look at this method with an example.

Example.

Given a matrix . Find the inverse matrix.

Solution.

Let's accept . Equality gives us three systems of linear inhomogeneous algebraic equations:

We will not describe the solution to these systems; if necessary, refer to the section solving systems of linear algebraic equations.

From the first system of equations we have, from the second - , from the third - . Therefore, the required inverse matrix has the form . We recommend checking it to make sure the result is correct.

Summarize.

We looked at the concept of an inverse matrix, its properties, and three methods for finding it.

Example of solutions using the inverse matrix method

Exercise 1. Solve SLAE using the inverse matrix method. 2 x 1 + 3x 2 + 3x 3 + x 4 = 1 3 x 1 + 5x 2 + 3x 3 + 2x 4 = 2 5 x 1 + 7x 2 + 6x 3 + 2x 4 = 3 4 x 1 + 4x 2 + 3x 3 + x 4 = 4

Beginning of the form

End of form

Solution. Let's write the matrix in the form: Vector B: B T = (1,2,3,4) Main determinant Minor for (1,1): = 5 (6 1-3 2)-7 (3 1-3 2)+4 ( 3 2-6 2) = -3 Minor for (2,1): = 3 (6 1-3 2)-7 (3 1-3 1)+4 (3 2-6 1) = 0 Minor for (3 ,1): = 3 (3 1-3 2)-5 (3 1-3 1)+4 (3 2-3 1) = 3 Minor for (4,1): = 3 (3 2-6 2) -5 (3 2-6 1)+7 (3 2-3 1) = 3 Determinant of minor ∆ = 2 (-3)-3 0+5 3-4 3 = -3

Transposed matrix Algebraic additions ∆ 1,1 = 5 (6 1-2 3)-3 (7 1-2 4)+2 (7 3-6 4) = -3 ∆ 1,2 = -3 (6 1-2 3) -3 (7 1-2 4)+1 (7 3-6 4) = 0 ∆ 1.3 = 3 (3 1-2 3)-3 (5 1-2 4)+1 (5 3-3 4 ) = 3 ∆ 1.4 = -3 (3 2-2 6)-3 (5 2-2 7)+1 (5 6-3 7) = -3 ∆ 2.1 = -3 (6 1-2 3)-3 (5 1-2 4)+2 (5 3-6 4) = 9 ∆ 2.2 = 2 (6 1-2 3)-3 (5 1-2 4)+1 (5 3- 6 4) = 0 ∆ 2.3 = -2 (3 1-2 3)-3 (3 1-2 4)+1 (3 3-3 4) = -6 ∆ 2.4 = 2 (3 2- 2 6)-3 (3 2-2 5)+1 (3 6-3 5) = 3 ∆ 3.1 = 3 (7 1-2 4)-5 (5 1-2 4)+2 (5 4 -7 4) = -4 ∆ 3.2 = -2 (7 1-2 4)-3 (5 1-2 4)+1 (5 4-7 4) = 1 ∆ 3.3 = 2 (5 1 -2 4)-3 (3 1-2 4)+1 (3 4-5 4) = 1 ∆ 3.4 = -2 (5 2-2 7)-3 (3 2-2 5)+1 ( 3 7-5 5) = 0 ∆ 4.1 = -3 (7 3-6 4)-5 (5 3-6 4)+3 (5 4-7 4) = -12 ∆ 4.2 = 2 ( 7 3-6 4)-3 (5 3-6 4)+3 (5 4-7 4) = -3 ∆ 4.3 = -2 (5 3-3 4)-3 (3 3-3 4) +3 (3 4-5 4) = 9 ∆ 4.4 = 2 (5 6-3 7)-3 (3 6-3 5)+3 (3 7-5 5) = -3 Inverse matrix Results vector X X = A -1 ∙ B X T = (2,-1,-0.33,1) x 1 = 2 x 2 = -1 x 3 = -0.33 x 4 = 1

see also solutions of SLAEs using the inverse matrix method online. To do this, enter your data and receive a solution with detailed comments.

Task 2. Write the system of equations in matrix form and solve it using the inverse matrix. Check the resulting solution. Solution:xml:xls

Example 2. Write the system of equations in matrix form and solve using the inverse matrix. Solution:xml:xls

Example. A system of three linear equations with three unknowns is given. Required: 1) find its solution using Cramer formulas; 2) write the system in matrix form and solve it using matrix calculus. Guidelines. After solving by Cramer's method, find the "Solving by inverse matrix method for source data" button. You will receive the appropriate solution. Thus, you will not have to fill in the data again. Solution. Let us denote by A the matrix of coefficients for unknowns; X - column matrix of unknowns; B - matrix-column of free members:

Vector B: B T =(4,-3,-3) Taking into account these notations, this system of equations takes the following matrix form: A*X = B. If matrix A is non-singular (its determinant is non-zero, then it has an inverse matrix A -1. Multiplying both sides of the equation by A -1, we get: A -1 *A*X = A -1 *B, A -1 *A=E. matrix notation of the solution to a system of linear equations. To find a solution to the system of equations, it is necessary to calculate the inverse matrix A -1. The system will have a solution if the determinant of the matrix A is nonzero. Let's find the main determinant. ∆=-1 (-2 (-1)-1 1)-3 (3 (-1)-1 0)+2 (3 1-(-2 0))=14 So, determinant 14 ≠ 0, so we continue solution. To do this, we find the inverse matrix through algebraic additions. Let us have a non-singular matrix A:

We calculate algebraic complements.

∆ 1,1 =(-2 (-1)-1 1)=1

∆ 1,2 =-(3 (-1)-0 1)=3

∆ 1,3 =(3 1-0 (-2))=3

∆ 2,1 =-(3 (-1)-1 2)=5

∆ 2,2 =(-1 (-1)-0 2)=1

∆ 2,3 =-(-1 1-0 3)=1

∆ 3,1 =(3 1-(-2 2))=7

∆ 3,2 =-(-1 1-3 2)=7

X T =(-1,1,2) x 1 = -14 / 14 =-1 x 2 = 14 / 14 =1 x 3 = 28 / 14 =2 Examination. -1 -1+3 1+0 2=4 3 -1+-2 1+1 2=-3 2 -1+1 1+-1 2=-3 doc:xml:xls Answer: -1,1,2.

Methods for finding the inverse matrix, . Consider a square matrix

Let us denote Δ =det A.

The square matrix A is called non-degenerate, or not special, if its determinant is nonzero, and degenerate, or special, IfΔ = 0.

A square matrix B is for a square matrix A of the same order if their product is A B = B A = E, where E is the identity matrix of the same order as the matrices A and B.

Theorem . In order for matrix A to have an inverse matrix, it is necessary and sufficient that its determinant be different from zero.

The inverse matrix of matrix A, denoted by A- 1, so B = A - 1 and is calculated by the formula

, (1)

where A i j are algebraic complements of elements a i j of matrix A..

Calculating A -1 using formula (1) for high-order matrices is very labor-intensive, so in practice it is convenient to find A -1 using the method of elementary transformations (ET). Any non-singular matrix A can be reduced to the identity matrix E by applying only the columns (or only the rows) to the identity matrix. If the transformations perfect over the matrix A are applied in the same order to the identity matrix E, the result will be an inverse matrix. It is convenient to perform EP on matrices A and E simultaneously, writing both matrices side by side through a line. Let us note once again that when searching for the canonical form of a matrix, in order to find it, you can use transformations of rows and columns. If you need to find the inverse of a matrix, you should use only rows or only columns during the transformation process.

Example 2.10. For matrix find A -1 .

Solution.First we find the determinant of matrix A
This means that the inverse matrix exists and we can find it using the formula: , where A i j (i,j=1,2,3) are algebraic additions of elements a i j of the original matrix.

Where .

Example 2.11. Using the method of elementary transformations, find A -1 for the matrix: A = .

Solution.We assign to the original matrix on the right an identity matrix of the same order: . Using elementary transformations of the columns, we will reduce the left “half” to the unit one, simultaneously performing exactly the same transformations on the right matrix.
To do this, swap the first and second columns: ~ . To the third column we add the first, and to the second - the first, multiplied by -2: . From the first column we subtract the second doubled, and from the third - the second multiplied by 6; . Let's add the third column to the first and second: . Multiply the last column by -1: . The square matrix obtained to the right of the vertical bar is the inverse matrix of the given matrix A. So,
.